Question: Let $p=x^2-7$. Which equation is equivalent to $(x^2-7)^2-4x^2+28=5$ in terms of $p$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $p^2-4p+23=0$ (Choice B) B $p^2+4p+23=0$ (Choice C) C $p^2-4p-5=0$ (Choice D) D $p^2+4p-5=0$
Solution: We are asked to rewrite the equation in terms of $p$, where ${p}={x^2-7}$. In order to do this, we need to find all of the places where the expression ${x^2-7}$ shows up in the equation, and then substitute ${p}$ wherever we see them! For instance, note that $-4x^2+28=-4({x^2-7})$. This means that we can rewrite the equation as: $(x^2-7)^2-4x^2+28=5$ $({x^2-7})^2-4({x^2-7})=5$ [What if I don't see this factorization?] Now we can substitute ${p}={x^2-7}$ : $({p})^2-4({p})=5$ Finally, let's manipulate this expression so that it shares the same form as the answer choices: ${p}^2-4{p}-5=0$ In conclusion, $p^2-4p-5=0$ is equivalent to the given equation when $p=x^2-7$.